0.2 m^(3) of an ideal gas at a pressure of 2 Mpa and 600 K is expanded isothermally to 5 times the initial volume. It is then cooled to 300 K at constant volume and then compressed back polystro-pically to its initial state. Determine the work done and heat transfer during the cycle.

0.2 m^(3) of an ideal gas at a pressure of 2 Mpa and 600 K is expanded

1.	0.2 m^(3) of an ideal gas at a pressure of 2 Mpa and 600 K is expanded isothermally to 5 times the initial volume. It is then cooled to 300 K at constant volume and then compressed back polystro-pically to its initial state. Determine the work done and heat transfer during the cycle.
Answer» Solution :Considering process 1-2 which is isothermal, i.e., `T_(1)=T_(2)=600K` ` P_(2)= (p_(1)V_(1))/V_(2)=(2xx10^(6)xx0.2)/(5xx0.2)=400xx10^(3) N//m^(2)=0.4 MPA` Work done, `W_(i-2)=p_(1)V_(1) log_(e). V_(2)/V_(1)=2xx10^(6)xx0.2 log_(e) 5 = 643775 J= 643.775 kJ` Considering process2-3 which is at constant volume, i.e., `V_(2)=V_(3)=5xx0.2=1 m^(3)` `P_(3)=P_(2)xxT_(3)/T_(2)=400xx10_(3)xx300/600 = 200xx10^(3) N//m^(2)=0.2 MPa` `W_(2-3)` = 0 as volume remains constant. For the polytropic process `3-1 =P_(3)V_(3)^(n)=p_(1)V_(1)^(n)` `n=(log_(e).(p_(1))/(p_(3)))/(log_(e). V_(3)/V_(1))=(log_(e). 2/0.2)/(log_(e). 1/0.2)=2.3026/1.6094=1.433` `W_(3-1)=(p_(3)V_(3)-p_(1)V_(1))/(n-1)=(0.2xx10^(6)xx1-2xx10^(6)xx0.2)/(1.433-1)` `=-461894 J=-461.894 kJ` Net work done, `W_(n)=W_(1-2)+W_(2-3)+W_(3-1)=643.774+0+(-461.894)=181.88 kJ` :. Heat transfer during the complete cycle=181.88 kJ